C++中char*与char(*)[]的区别及shared_ptr编译报错咨询

Let's walk through your problem clearly—array pointers in C++ can be tricky when you're just starting out, so it's totally normal to hit this confusion.

First, Fixing Your Code #

The immediate issue is that you're overcomplicating the assignment. When you use shared_ptr<char[]>, its get() method returns a char* (a pointer to the first element of the char array), not a pointer to the entire array (char(*)[]). You don't need to dereference it at all:

shared_ptr<char[]> var(new char[20]);
char *varptr = var.get(); // This works perfectly

The error you saw happened because *(var.get()) tries to dereference a char*, which gives you a single char value—not an array. Assigning a char to a char* is invalid, hence the compiler's complaint.

Explaining char* vs char(*)[] #

Let's break down the two pointer types you're confused about:

1. char* #

• This is a pointer to a single char value.
• In practice, it's commonly used to point to the first element of a char array (since array names implicitly convert to pointers to their first element).
• Example:

  char arr[20] = "hello";
  char* p = arr; // p points to arr[0] (the 'h')
  

2. char(*)[] #

• This is a pointer to an array of chars (specifically, a pointer to an unknown-size char array). For a fixed-size array, it would look like char(*)[20] (pointer to a 20-element char array).
• This type points to the entire array as a single object, not just its first element.
• Example:

  char arr[20] = "hello";
  char(*p_arr)[] = &arr; // p_arr points to the entire array
  

Can char(*)[] Be Assigned to char*? #

Yes, but you need to dereference the array pointer first to get the array (which then "decays" to a pointer to its first element):

char arr[20] = "hello";
char(*p_arr)[] = &arr;
char* p = *p_arr; // *p_arr is the array, which decays to char* pointing to arr[0]

But in your original code, this step wasn't necessary because shared_ptr<char[]>::get() already returns the char* you need directly.

Why Your Initial Assumption Was Off #

You looked at the general shared_ptr::get() declaration T* get() const noexcept;, and assumed that for shared_ptr<char[]>, T is char[], so T* would be char(*)[]. However, shared_ptr has a specialization for array types (shared_ptr<T[]>) that adjusts the return type of get() to T*—which for T=char is char*, not a pointer to the array. This specialization exists to make working with array smart pointers more intuitive, matching how raw array pointers behave in most cases.

内容的提问来源于stack exchange，提问作者Ojs